Question

The probability that a student correctly answers on the first try (the event

Answer 1

Note that the two events are mutually exclusive. If the question is answered correctly on the first try, there's no need to give it another attempt. So $\mathbb P(A\cap B)=0$.

We're given that $P(A)=0.2$ and $P(B\mid A^C)=0.5$. From the first probability, we know that $P(A^C)=1-0.2=0.8$. By definition of conditional probability,


$\mathbb P(B\mid A^C)=\dfrac{\mathbb P(B\cap A^C)}{\mathbb P(A^C)}$
$\implies\mathbb P(B\cap A^C)=0.5\cdot0.8=0.4$

We're interested in the probability of either $A$ or $B$ occurring, i.e. $\mathbb P(A\cup B)$. Apply the inclusion-exclusion principle, which says

$\mathbb P(A\cup B)=\mathbb P(A)+\mathbb P(B)-\mathbb P(A\cap B)$

We know the probability of intersection is 0, and we know $\mathbb P(A)$. Meanwhile, by the law of total probability, we have

$\mathbb P(B)=\mathbb P(B\cap A)+\mathbb P(B\cap A^C)=\mathbb P(B\cap A^C)$

so we end up with

$\mathbb P(A\cup B)=0.2+0.4=0.6$